 So, let us continue our discussion with the density operator treatment of NMR experiments. We have described previously the formal definitions of the density operator, the time evolution of the density operator and we have shown what kind of an equation governs the time evolution of the density operator. We have also seen what kind of a solutions exist in the presence of the RF and in the absence of the RF. So, a generalized NMR experiment which will consist of several pulses with interleaving evolutions under the influence of different Hamiltonians and this is as indicated here in this chart. This is the repetition from the previous one for continuity. So, you have you start an experiment here with the density operator represent as 0, 0 and then you apply a pulse P1. It may be any kind of a pulse may be a 90 degree pulse, 180 pulse, 45 degree pulse or whatever that is up to one's choice. Then this is followed by an evolution period, free precision period which is under the influence of the Hamiltonian H1 and this happens for the time tau 1. Then this is followed by another pulse we call it as P2 and once again it is followed by evolution under the influence of a different Hamiltonian H2 and that happens for a time tau 2 and likewise you continue for the P3 pulse then you have the evolution H3 tau 3, P4 pulse and so on and so forth. Till you complete all your pulses and you collect the data sometime here and that is where the information is analyzed subsequently. So, you start from some particular time point here which call it as 0 and you collect the data here with various kinds of manipulations happening in the various time periods here. The data you collect will carry all the information about that has happened in this previous times and therefore we have to calculate this. So, we must know how to calculate the evolution of the spin system under the influence of all of this and we have already shown that such a generalized pulse sequence can be represented by a time evolution of the density operator in this manner. You start with rho of 0 and you apply a pulse P1. The transformation happens as P1 rho 0 P1 minus 1 and this density operator now becomes the time t is equal to 0 for this portion of the evolution and then it continues here then that is represented by the e to the minus i by h cross H1 tau 1 and e to the i by h cross H1 tau 1. Now, this whole thing becomes the time t is equal to 0 so far as the pulse P2 is concerned and then this is how it continues. So, you see here as you are progressing in the calculation we go from the laboratory frame here to the interaction representation here normal representation interaction representation back to the laboratory representation again interaction representation here back to the normal representation and so on. So, keep switching between the two representations for the calculation purposes and these two representations are generally separated out in this way. You see this whole calculation consists of these two kinds of evolutions. One is the e to the minus i by h cross ht rho e to the i by h cross ht this is free evolution this will apply to this period this period and so on and for the pulses we have this sort of a transformation P rho P minus 1 and this will apply to this this one this one this one this one and so on. So, this is a quite a involved calculation as you could as we saw earlier because each one of these operators will have to be represented as a matrix and the matrix is become larger and larger as we go to larger and larger spin systems and different kinds of flip angles of the pulses different phases of the pulses all of that will complicate the matters quite substantially. So, therefore a simplification was developed and that is called as the product operator formalism. To simplify the above calculation the product operator formalism has been developed for weakly coupled spin systems. Notice here the key elements key points here I want to like to underline here one is this weakly coupled spin systems you have to highlight this point this is applicable to weakly coupled spin systems. The density operator is expressed as a linear combination of some basis operators which are represented as BS and these constitute a complete set. So, these are important terms here which one has to understand. You have already studied these situations earlier but let me try and recap these to some extent. What is a weakly coupled spin system? So, weakly coupled spin system this was an important thing to remember. So, if you have two spins both are spin half and if you have a coupling constant between them J and the chemical shift separation between them is delta then if it is much less than 1 then we say it is a weakly coupled spin system and this has been studied earlier when we actually went through the analysis of NMR spectra how the spectra get complicated when this condition is not satisfied. By and large in the modern spectrometers when you have very high magnetic fields this condition is usually satisfied for most systems and therefore it is convenient to work with the theories which are developed for weakly coupled spin systems. And what is the consequence of this? So, if you look at the spin systems and you want to calculate the various eigenstates and the energy levels of this it turns out that the product functions are the eigenfunctions. If I for example one spin the states are alpha and beta this is one spin. For two spins each one of them is alpha beta then we said for we have the product functions alpha alpha alpha beta alpha and beta beta these are the eigenfunctions of the spin system. For three spin systems then we will have product of three functions we have alpha alpha alpha alpha alpha alpha beta alpha beta alpha beta beta alpha and then alpha alpha beta alpha beta beta beta alpha beta beta beta beta beta beta. So, we have eight eigenfunctions here. So, the reason why I am saying is the product functions are the basis set. So, here all these functions represent the complete spin system. Any situation can be expressed as the linear combination of these wave functions. So, that takes me to the next point there just as you are able to write a basis set and able to describe the complete wave function for a given spin system as a linear combination of these various functions. If I call this as Un each one of them is called as Un then we wrote psi is equal to summation Cn Un we wrote like that earlier. So, these Un are this various basis functions and one can write a generalized wave function as a linear combination of these basis functions. The same principle will be used for the product operator formalism where we will represent the density operator as a linear combination of some basis operators. So, this is therefore we write rho of t as summation Bst this capital Bs these constitute what are called as the basis operators and these are the coefficients which will contribute to the summation here and that whatever are the ones which are contributing here that will determine the nature of the density operator at any point in time. So, what are these basis operators and these must form a complete set just as we talked about the Eigen functions the forming a complete basis that is so far as the wave function is concerned here these operators must form a complete basis set so that we can represent any density operator as a linear combination of this. Now having said that now let us try and calculate what is the transformation that happens for these density operators for the rho dash now we introduced the rho of t here the rho of t was here so therefore now it becomes summation Bs e to the power minus i by h cross ht Bs e to the i by h cross ht ht and similarly for the pulses we have the rho double dash which is summation Bs p Bs p minus 1. Now how do we choose these basis operators there can be there are many ways of choosing these basis operators and we will take only one of those examples which is the one which is most commonly used for various calculations. For certain situations some other kinds of basis operators sets are also used but we will use the one particular one which is most commonly used in every case however these are formed from what are what we know already as angular momentum operators these are formed from angular momentum operators which is a very natural choice to make because your Hamiltonians consist of the angular momentum operators therefore there will be kind of operationally easier things to do when you have the basis operators also express in terms of angular momentum operators which do satisfy the condition of completeness and therefore the calculations also become relatively easy. Now one of the angular momentum operators typically we have in the Cartesian space this is the unit matrix and we have the three angular momentum operators ix, iy and iz so we build on this we build using this so for a single spin if it is then I will have one ix, iy and iz and e is a unit operator and this is the normalization factor half which is there in every case. Now if I have n spins so we said for one spin we had four operators right one was the unit matrix and then you have ix, iy and iz so if I have n spins I must I will have how many of how many combinations I can make from each one of those each one of them has four possibilities and therefore I can create 4 to the power n combinations of these various kinds of products I can take because I should not be missing any portion of the density operator so all the spins must be represented all situations of all the spins must be represented therefore all the operators of the individual spins must find a representation in your basis set. Therefore if I have n spins in a coupled network there will be total of 4 to the power n elements in the basis operator sets so for example for two spins there will be total of 16 operators so Cartesian space these are we have this e dash this is basically related to the unit matrix we have this one half that one we will explicitly see what those ones are then you have the x, y, z of one spin and then the x, y, z of the second spin then I take individual products of each one of these with the other spin and this is again a normalization factor here too I will add 2 1x 2x 1x 2y 1x 2z and next 1y 2x 1y 2y 1y 2z and 1z 2x 1z 2y 1z 2z and these are the normalization factors in each of these cases. So likewise if you go for 3 spins how many operators you will have notice in the previous case we had a total of 16 4 to the power 2 now here how many we should have if you have 3 spins I should have 4 to the power 3 though that means 64 operators so here it is the unit matrix once more and then I will have here I a, p, I m, p, I q, p where p goes from x, y, z so there will be total of 9 operators a, m and q are the 3 spin symbols and p it goes from x, y, z this is the Cartesian index. So and then I will have 2 spin products so 2 I a, p, I m, r, 2 I m, p, I q, r and then 2 I a, p, I q, r and here notice I am varying a, m, p and r both so there are a, m 3 spins a, m, q right so I will have product a to m, m to q, a to q these 3 products will be there and for each case I will have the p and r varying as x, y, z so there will be total how many total of 27 operators so this is there are 3 here p takes 3 values r takes 3 values so therefore 3 into 3 into 3 so you will have total of 27 operators and then you will have the 3 spin products because all the 3 spins combinations can be present so you have 3 spin products so here again you will have a normalization factor 4 now I have the products of all the 3 spin operators here corresponding to each of the 3 spins so a, m, q and then I will have the here indices p, r, s all the 3 p, r, s can go from x, y, z so therefore this is 3 to the power 3 and I will have again 27 operators here so total of 64 operators I get for the 3 spin system so likewise I will get many more as I go to larger and larger spin systems however by and large we generally do not come across operators which are more than 3 or 4 now let us see what do these operators encode we created the various basis operators and we said if they represent the density operator and we have seen the density operator contains various kinds of populations single quantum coherence is double quantum coherence is zero quantum multiple quantum various kinds of coherence is present now what do these base operators represent which base operators this is basis operators which basis operator represents what sort of information what kind of information does it contain so now for one spin the Cartesian representations in the term of matrices are as you have seen here ix is with this we have already calculated previously is half 0 1 1 0 so there are two states alpha beta alpha beta and so therefore this is an off diagonal element in the if you want to take this matrix and if you put this in the density operator this off diagonal element will imply a single quantum coherence and that means it is x magnetization of a particular spin this is the one spin so therefore in the case of one spin it is the x magnetization of that spin now i y is half 0 minus y i 0 this is once again a single quantum coherence because this is off diagonal element here states alpha alpha beta alpha and beta alpha and beta so this is a transition from alpha to beta therefore single quantum coherence here but this is a y magnetization because you have see your imaginative quantity here the i z here is the diagonal elements are non-zero off diagonal elements are 0 therefore this represents the populations and the populations will indicate to you the z magnetization okay now when we go to two spins we calculate the matrix representations of the two spin operators in the following manner we follow the same trick as we did for for the pulses you notice here I have here the unit operator as e dash and e dash is calculated simply as this is the unit matrix here and this is again a unit matrix you take a direct product because for two spins k and l it will be a 4 by 4 matrix and where what does this represent so I take a direct product here 1001 into 1001 so I will get here is not written here but this will be only diagonal elements will be present we can write that explicitly we can write here this will be 1111 and all others will be 0 right 1001 this will be 0 this will be 0 1001 and that is the direct product if you take kx now in a two spin system I am trying to represent the k magnetization x magnetization of the k spin alone so therefore my operator for k spin is 0110x magnetization then I take a direct product with a unit matrix here so that is how I get a 4 by 4 matrix so if I take this direct product here 0 into this matrix give me this 0 matrix here and 1 into this matrix give me the same matrix 1001 and similarly here this 0 into this 1 into this matrix give me 1001 and 0 into this gives me 000 so therefore now what do I get here now this is the if this way the density operator you see what this represents this represents the 1 3 coherence we will show the explicitly energy level diagram the next slide this is the 1 3 coherence this is the 2 4 coherence and these are the other corresponding elements on the other side so therefore this represents x magnetization take it as the k spin and these are in phase single quantum coherences seen phase because these two coherences have the same sign if I do ilx now this has to this is another spin so how do I get this to get this I will have to simply reverse this order unit matrix I put it here and the x operator I put it here so therefore unit matrix come 1001 and 0110 if I now take the direct product of this you see I get these elements which are non-zero in this case we had these elements non-zero here I will get these elements non-zero and all other elements will be zero so this will be now 1 2 2 1 1 2 and 3 4 again 2 coherences which obviously belong to the different spin and that we will see in the next slide here so if I take kx I put the same matrix here and you see this is the 1 3 coherence and these are the energy levels of the 2 spins 1 2 3 4 this is the 1 3 transition right so 1 3 transition is this 1 3 coherence is this and here it is a k spin which is flipping l spin is the same alpha l and it is the k spin which is going from alpha to the beta and similarly here in the 2 4 transition it is the k spin which is going from alpha to the beta l spin remains the same therefore these 2 are positive these therefore this represent k spin coherences and therefore that represent the k spin magnetization x spin now if I were to draw this in terms of the coherence as the NMR spectrum because the coherence will appear as magnetization which is observable in the NMR spectrum so therefore these 2 transitions will be represented as 1 3 2 4 and both have the same sign because these coherences have the same sign now if I take the other one I lx these elements are non-zero now what are these this is 1 2 1 2 is this and 3 4 is this okay now here it is the l spin which is flipping the k spin is constant that is why these are the l spin coherences okay so once again therefore this implies x magnetization of the l spin which is in phase single quantum coherence what it means in phase they have the same sign therefore we call them as in phase and these 2 transitions are in the NMR spectrum will appear in this manner 1 2 and 3 4 now ik y I do the same thing I have this ky operator here and the unit matrix here and therefore I get here the y magnet the same elements except that I get here minus i minus i i i and similarly for l I get here minus i i here and minus i i here so this represents the l spin magnetization this represents k magnetization but what about this i i means it is imaginary so this is 90 degree is out of phase and that we call as the dispersive line shape this to represent a dispersive line shape because if I were to measure the x magnetization as in phase then the y magnetization will be the 90 degree out of phase the 90 degree out of phase is the dispersive line shape so and how does that look so here we have here the ik y these are the 2 transitions here the same here 1 3 2 4 but in the NMR spectrum we will look at the 90 degrees phase shift this will be a dispersive line shape but these are in phase in phase meaning this goes up and down here and this also goes up and down in this manner so therefore these 2 transitions are in phase these are dispersive in phase transitions similarly for ly we have 1 2 and 3 4 and these are once again in phase transitions and 1 2 3 4 and that is because I have the same sign for this and this so for this is clear about the 2 spin system how the individual elements of the base operators are to be interpreted what is the information content in the different elements of the base operators what do they represent now extended to 3 spin the 3 spin we will make a beginning here of course we will continue later with more details continue take a 3 spin system which is KLM these are the 3 spins so now I said there earlier that there will be total of 64 operators we will illustrate only some of those here we take ikx ilx imx which are supposed to represent the x mechanization of KL and M now each one of them is will have how many transitions there are suppose there is a coupling between k and l and l and m and k and m all the 3 spins are coupled then I will have 4 transitions for each one of those because each one of them leads to a splitting into 2 therefore I will have 4 transitions now the energy level diagram for the 3 spin system there will be 8 energy levels as I mentioned to you earlier the case of 2 spins here I will have alpha k alpha l ilm alpha m see essentially 1 2 spin system another 2 spin system except now you add the 3rd spin as alpha here and the beta here so you generate a 8 level energy level diagram so now each one of them now the duplication happens this states belong to one this transitions belong to one spin all the red ones belong to one spin and this green ones belong to one spin as before and now we have the 3rd spin transitions happening from here to here here to here here to here and here to here and all of those are represented in this manner since all of them are in phase I must write them as absorptive line shapes here for each one of those individual transitions are indicated here for the l magnetization the k magnetization and the m magnetization so therefore this is the simplest here we are seeing the individual transitions of the k and l spins and this we call them as in phase magnetization because they are in phase with respect to the orientation of the other 2 spins so it regardless of what the other 2 spins are other 2 spin states are I have the same positive absorption absorption signal therefore we call them as in phase the other things we will see in the next class we stop here